Get employees who received a raise in 2 consecutive years - mysql

Question

I am trying to Get employees who received a raise in 2 consecutive years, in this case employee 1000 is the right answer.
here is the data and the sql i have tried.
EID
SALARY
YEAR
1000
10,000
2016
1000
7,500
2015
1000
6,000
2014
1001
8,000
2016
1001
7,500
2015
1002
7,500
2016
1002
7,500
2015
1002
5,000
2014
1003
6,000
2016
1003
7,000
2015
1003
5,000
2014
i have used following code however it gets only row number by EID and not calcualtion of last year and present year, i need to find employee who got raise in 2 consecutive years.
output
select * ,
row_number() over(partition by eid order by salary and year desc)as rn
from gs;

Answer 1

You can do it using the LEAD window function that compute the two consecutive previous value of the salary. Then you can check how many employees have at least one row with salary1 < salary2 < salary3.
SELECT DISTINCT
eid
FROM (
SELECT
eid,
year,
salary,
(LEAD(salary, 1) OVER(PARTITION BY eid ORDER BY year DESC)) AS prev_salary1,
(LEAD(salary, 2) OVER(PARTITION BY eid ORDER BY year DESC)) AS prev_salary2
FROM
employees
) consecutive3
WHERE
salary > prev_salary1
AND
prev_salary1 > prev_salary2
The assumption is that there are no missing years for which a salary of a dependent was not recorded.
Here's the fiddle: https://dbfiddle.uk/?rdbms=mysql_8.0&fiddle=8c0d8a1deec8e77bb32a173656c3e386.
EDIT: Detailed explanation
Let's do the example of Jennifer, who has worked for five years and got these salaries:
2018 -> 65000
2017 -> 55000
2016 -> 50000
She's a candidate for being selected as her salary was raised three times consecutively.
1. LEAD(salary, 1) OVER(PARTITION BY eid ORDER BY year DESC)
Allows you to get the salary for year "Y" and the salary for year "Y-1":
("year" -> "salary", "previous_salary")
2018 -> 65000 , 55000
2017 -> 55000 , 50000
2016 -> 50000 , NULL
2. LEAD(salary, 2) OVER(PARTITION BY eid ORDER BY year DESC)
Allows you to get the salary for year "Y" and the salary for year "Y-1":
("year" -> "salary", "previous_salary", "previous_salary_by_2_years")
2018 -> 65000 , 55000 , 50000
2017 -> 55000 , 50000 , NULL
2016 -> 50000 , NULL , NULL
3. WHERE salary > prev_salary1 AND prev_salary1 > prev_salary2
Some filtering on the employees who
have their year3 salary higher than their year2 salary (salary > prev_salary1)
have their year2 salary higher than their year1 salary (prev_salary1 > prev_salary2)

Answer 2

I know that this has already been answered but here is my take using the lag function to determine if there was an increase from the previous year and ran that twice.
SELECT *
FROM (
SELECT
t2.*,
LAG(increase) over (partition by eid order by year) AS increaseNextYear
FROM (
SELECT
t1.*,
COALESCE(salary - LAG(salary) over (partition by eid order by year), 0) > 0 AS increase
FROM tbl_test t1
) as t2
) t3 where increase AND increaseNextYear

Answer 3

with
dates as
(
select
a.*,
dense_rank() OVER (
partition by eid
order by year desc, salary
)
as rn,
case
when
lead(salary,2)over(partition by eid order by year, salary)
>
lead(salary,1)over(partition by eid order by year, salary)
and
lead(salary,1)over(partition by eid order by year, salary)
>
salary
then
1
else
0
end
as flag
from
employees a
)
select
eid
from
dates
where
rn = 3
and flag = 1

Answer 4

Not a beautiful query, but straight-forward: find employees who had a salary in a year where the salary in the previous year was lower and the salary in the year before that even lower. Using LAG is more elegant, but I thought I'd throw this in, just to show an alternative.
select *
from employee
where exists
(
select null
from gs
where gs.eid = employee.id
and exists
(
select null
from gs prev
where prev.eid = gs.eid
and prev.year = gs.year - 1
and prev.salary < gs.salary
and exists
(
select null
from gs prevprev
where prevprev.eid = prev.eid
and prevprev.year = prev.year - 1
and prevprev.salary < prev.salary
)
)
);
Same thing with a join:
select *
from employee
where exists
(
select null
from gs
join gs prev on prev.eid = gs.eid
and prev.year = gs.year - 1
and prev.salary < gs.salary
join gs prevprev on prevprev.eid = prev.eid
and prevprev.year = prev.year - 1
and prevprev.salary < prev.salary
where gs.eid = employee.id
);

Answer 5

For versions prior to 8.0 (mine is 5.7) which lack the cutting edge features of the newer one, I tried a procedure to accomplish the job. First and foremost, get all the eid which have no less than three years' salary record, which is the minimum requirement of the consecutive bonus. The rest is to fetch and compare using a cursor from the eid pool. The result is stored in a temporary table t .
delimiter //
drop procedure if exists lucky_emp//
create procedure lucky_emp()
begin
declare old_eid int default 0;
declare old_salary int;
declare new_eid int ;
declare new_salary int;
declare bonus_year int;
declare fin bool default false;
declare c cursor for select eid,salary from salary where eid in(select eid from salary group by eid having count(eid)>=3) order by eid,year;
declare continue handler for not found set fin=true;
drop temporary table if exists t ;
create temporary table t (t_eid int);
open c;
lp:loop
fetch c into new_eid ,new_salary;
if fin=true then
leave lp;
end if;
if new_eid !=old_eid then
set old_eid=new_eid,old_salary=0,bonus_year=0;
end if;
if new_salary> old_salary then
set bonus_year=bonus_year+1,old_salary=new_salary;
else
set bonus_year=0;
end if;
if bonus_year=3 and new_eid not in(select t_eid from t) then
insert t values(new_eid);
end if;
end loop lp;
end//
delimiter ;
select * from t ;

Answer 6

Select a.*, b.prev_sal1, b.prev_sal2
from employees a
join (
Select eid ,year,
lag(salary,1) over (partition by eid order by year) as prev_sal1,
lag(salary,2) over (partition by eid order by year) as prev_sal2
from employees ) b
on a.eid=b.eid
and a.year = b.year
where salary>prev_sal1 and prev_sal1>prev_sal2
fiddle: https://dbfiddle.uk/rfGv31zM